With the k-th appended term being 2*3*...*(2+k-2)*2^k*(2^k-1)*Bern(k) / (2*k!*(J^(k+2-1))). Bern(k) is a Bernoulli number and J is a large number of the form 4n + 1. See equation 3:3:7 in Spanier and Oldham. - Harry J. Smith, May 07 2009

c = (Zeta(2, 1/4) - Pi^2)/8 = (Psi(1, 1/4) - Pi^2)/8, with the Hurwitz Zeta function and the trigamma function Psi(1, z). For the partial sums of the series given in the name see A294970/A294971. - Wolfdieter Lang, Nov 15 2017